In a radioactive material, fraction of active | vedclass

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Question

First order decay

$N ( t )= N _{0} e ^{-\lambda 1}$

Given $N ( t ) / N _{0}=9 / 16= e ^{-\lambda}$

Now, $N ( t / 2)= N _{0} e ^{-2} v / 2$

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

First order decay

$N ( t )= N _{0} e ^{-\lambda 1}$

Given $N ( t ) / N _{0}=9 / 16= e ^{-\lambda}$

Now, $N ( t / 2)= N _{0} e ^{-2} v / 2$

$\frac{ N ( t / 2)}{ N _{0}}=\sqrt{ e ^{-\lambda t}}=\sqrt{9 / 16}$

$N(t/2)=3/4 N _{0}$

In a radioactive material, fraction of active material remaining after time $t$ is $\frac{9}{16}$ The fraction that was remaining after $\frac{t}{2}$ is

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